Suppose you system of units has one years as the unit of tiem , one light year as the unit of length and mass of earth as the unit of mass. How much 1 N be equal to the units of force in this system. Take 1 year = `3 xx 10^(5) s , ` 1 light year `= 9.5 xx 10^(15)` m and mass and earth `= 6 xx 10^(24)` kg .

Explanation

Let y denote year, 1 denotes light year and `m_(e)` mass of earth . `n_(1)U_(1) =n_(2)U_(2)`
`:. N_(2)=n_(1) (U_(1)/U_(2))`
1 N is the unit of force and the dimensions of force are `MLT^(-2)`.
`:. N_(2) = n_(1) [M_(1)/M_(2)] [L_(1)/L_(2)] [T_(1)/T_(2)]^(-2) `
Here subscript 1 represents the system of units in which .1N. is the unit of force i.e, SI, and subscritpt 2 represents the new system.
`:. n_(2)=1xx[(1kg)/(1m_(e))] [(1m)/(1l)][(1s)/(1y)]^(-2) =[(1kg)/(6xx10^(24)kg)] [(1m)/(9.5xx10^(15)m)][(1s)/(3xx10^(5)s)]^(-2)`
`=[(3xx10^(5))^(2)/(6x10^(24)xx9.5xx10^(15))] =(9xx10^(10))/(6xx9.5xx10^(39))=1.58xx10^(-30)`
`In=1.58xx10^(-30)` units of force in the new system . `:. IN = 1.58xx10^(-30) m_(e)1 y^(-2)`